The Linear Pendulum

We can represent a linear pendulum by the differential equation

$$\ddot{x}=-\omega ^{2}x \Rightarrow X=\begin{bmatrix} x\\ \dot{x} \end{bmatrix}, \dot{X}=\begin{bmatrix} 0 & 1 \\ -\omega ^{2} & 0 \end{bmatrix} X \,$$

We can use a change of variables such that

$$\begin{cases} x_{1}=-\omega x \\ x_{2}=\dot{x} \end{cases} \Rightarrow \dot{X}=\begin{bmatrix} 0 & -\omega \\ \omega & 0 \end{bmatrix} \begin{bmatrix} x_{1}\\ x_{2}\end{bmatrix}\,$$

which is easily solvable using what we have learned.

If there is friction, the equation changes slightly, and we have

$$\ddot{x}=-\omega ^{2}x - \lambda \dot{x} \Rightarrow \dot{X}=\begin{bmatrix} 0 & 1 \\ -\omega^2 & -\lambda \end{bmatrix} X \Rightarrow \begin{cases} T = -\lambda \\ \Delta = \omega^2 \end{cases} \,$$

Therefore, if $$\lambda < 4 \omega^2$$ we have a focus, otherwise we have a node.